What makes 37

Example 1: Stuart is a carpenter. He can build 1 wardrobe in 3 days. How long will it take him to make 37 wardrobes?

Stuart makes 1 wardrobe in 3 days. Example 2: Lucas owns a fruit shop. Three of his customers ordered 6 mangoes, 12 mangoes, and 18 mangoes respectively.

Help Lucas find whether the mangoes in total can be grouped in 37s. If this cannot be done, then find how many more mangoes are required to be added to group them as 37s? The factors of 37 are 1, 37, and factors of 38 are 1, 2, 19, The factors of 37 are 1, 37 and the factors of 21 are 1, 3, 7, This implies that 37 and 21 are co-prime. Since the factors of 37 are 1, 37 and factors of 20 are 1, 2, 4, 5, 10, Hello, from Mr.

Fearn who teachers at East Park Primary School who sent in at a much later date the following:- I'm a regular user of your fabulous range of problems with my Y5 and 6 classes. This morning, we had a go at 'Make 37' and had a great time with the investigation.

I was very impressed with their ingenuity! I am also impressed. I wonder if there are more solution of this kind? The following three submissions are from Michelle L, Michelle D and Cathy who go to Greenacre Public School Australia: The problem 'Make 37' requires you to use ten of any of the numbers 1, 3, 5, 7 to make a total of Note that the numbers are all odd and 37 is also an odd number, but you need ten of the odd numbers to make another odd number.

Usually, an odd number added to another odd number equals an even number. Three odd numbers added together equals an odd number. Four odd numbers added together equals an even number. Five odd numbers added together equals an odd number. Six odd numbers added together equals an even number and so on. When you get up to ten numbers, you will find that the solution is an even number. You can make 37 with nine of those numbers or eleven of them because nine or eleven odd numbers added together equals an odd number.

The solution for "Make 37" is that it is impossible to make 37 with the numbers 1, 3, 5 and 7 if you add them ten times. Bit shifting, while being a Java operator, is also not a math operator in the true sense of the word human beings have no need to bit encode digits to manipulate them, that's a computer thing.

You're best bet would be to stick with the normal mathematical operators imho. Michael Dunn. Mark Herschberg. Which of the following is allowed: 1 Factorial Most problems of this nature do accept! Those who do not argue that you need two numbers to define a summation. Those who do, allow a single number, with the terminal base to be implied as zero.

Are other roots? A root sign typically has a a number associated with it. Some allow for a root sign sans number to imply a square root. Well, okay. So is factorial! In that case Michael Dunn was pretty close. I'm sure this works better in a non-programmer crowd.

In other words, pretty much the basic stuff. No functions floor, ceiling, rounding, trig functions, etc No increment or decrement pre or post. I suppose square root would probably be allowed, but I'm guessing not other roots as that would call for an additional number.

In any case the solution does not call for it. Nor does the solution call for raising anything to the 5th or 55th or th power not sure if that is even acceptable in these types of problems though.